I recently learned the following pleasant fact. (It was in the proof of Proposition 3.1 of this paper - but don't worry, there's no model theory in this question.)

Let $G$ be a group, and let $a,b,c\in G$. If $aba^{-1} = b^2$, $bcb^{-1} = c^2$, and $cac^{-1} = a^2$, then $a = b = c = e$. Put another way, the group defined by generators and relations $\langle a,b,c \mid aba^{-1} = b^2, bcb^{-1} = c^2, cac^{-1} = a^2 \rangle$ is the trivial group.

I came up with the following elementary, but ugly, proof:

The relations can be rewritten as (1) $ab = b^2a$, (2) $bc = c^2b$, (3) $ca = a^2c$.

Using (1), (2), and (3), we can rewrite $a^4bc = a^4c^2b = c^2ab = c^2b^2a$.

But we can also rewrite $a^4bc = b^{16}a^4c = b^{16}ca^2 = c^{2^{16}}b^{16}a^2$.

So $c^{2^{16}}b^{16}a^2 = c^2b^2a$. This implies $a = b^{-16}c^{2-2^{16}}b^2$.

Substituting for $a$ in (1) above, $b^{-16}c^{2(1-2^{15})}b^3 = b^{-14}c^{2(1-2^{15})}b^2$, and cancelling from both sides, $c^{2(1-2^{15})}b = b^{2}c^{2(1-2^{15})}$.

But now by (2), we have $bc^{1-2^{15}} = b^{2}c^{2(1-2^{15})}$, and $b = c^{2^{15}-1}$. But then $b$ and $c$ commute, so $bcb^{-1} = c^2$ implies $c = c^2$, and $c = e$. It then follows easily that $a = b = c = e$.

Question: Is there a better way to see this? i.e. a more abstract proof, or at least one that doesn't involve manipulating words of length $2^{16}$?

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    $\begingroup$ A slightly relevant remark, if there are four generators with similar relations, the group will be infinite. That is, $⟨a,b,c,d∣aba^{−1}=b^2, bcb^{−1}=c^2,cdc^{−1}=d^2, dad^{−1}=a^2⟩$ is not finite. c.f. Serre's Trees Page 9. $\endgroup$
    – userabc
    Jan 4, 2019 at 5:02
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    $\begingroup$ ... and now that I can search for an exercise number in a well-known book, I find that there are a number of questions on this site about the same group. For example, Jim Belk gave almost an identical proof to the one I found here, and Martin Brandenburg asked essentially the same question I'm asking here. Together, this evidence makes me think that there's unlikely to be a nicer proof. It would be reasonable to close this question as a duplicate. $\endgroup$ Jan 4, 2019 at 5:26
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    $\begingroup$ Once you have proved that $a \in \langle b,c \rangle$, you can use the more conceptual argument given by Bhaskar Vashishth in the linked proof, that $G$ is perfect, but $G = \langle b,c \rangle$ is solvable, so $G$ is trivial. $\endgroup$
    – Derek Holt
    Jan 4, 2019 at 15:05
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    $\begingroup$ A quantitative question would go as follows: triviality of $G$ means that one can write, in $F_3$, $a=w$, $w=\prod_{i=1}^ng_ir_ig_i^{-1}$ with $r_i$ among the the three relators. (a) what's the smallest $n$ for this this exists (this is the "area" of the relation)? (c) what is the smallest length $\sum_i(2|g_i|+|r_i|)$ of such a writing? (c) what's the smallest radius (this is, in the loop given by $a^{-1}w$, the largest radius). $\endgroup$
    – YCor
    Jan 11, 2019 at 2:49
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    $\begingroup$ Does this answer your question? Presentation $\langle x,y,z\mid xyx^{-1}y^{-2},yzy^{-1}z^{-2},zxz^{-1}x^{-2}\rangle$ of group equal to trivial group $\endgroup$
    – mathlander
    Oct 5, 2022 at 14:47

1 Answer 1


As pointed out by Derek Holt in a comment, if we can show that $c\in \langle a,b\rangle_G$, then we can finish the proof in a "conceptual" way:

  • Since $a = [c,a]$, $b = [a,b]$, and $c = [b,c]$, $G' = G$, i.e., $G$ is perfect.
  • Since $c\in \langle a,b\rangle_G$, $G = \langle a,b\rangle_G$, and since $aba^{-1} = b^2$, $G$ admits a surjective homomorphism from $H = \langle x,y\mid xyx^{-1} = y^2\rangle$.
  • This group $H$ is known as the Baumslag-Solitar group $\mathrm{BS}(1,2)$, and is well-known to be solvable. One way to see this is to show that $H\cong \mathbb{Z}[\frac{1}{2}]\rtimes \mathbb{Z}$, where $\mathbb{Z}[\frac{1}{2}]$ is the additive group of dyadic rationals and $n\in \mathbb{Z}$ acts on $\mathbb{Z}[\frac{1}{2}]$ by multiplication by $2^n$. As a semidirect product of two abelian groups, $H$ is solvable.
  • As a quotient of the solvable group $H$, $G$ is solvable. But a perfect solvable group must be trivial.

I don't expect to find a "conceptual" proof that $c\in \langle a,b\rangle_G$. But I recently learned a beautiful visual proof of this fact due to Josh Hinman (shared here with his permission).

The relations $ab = b^2a$, $bc = c^2b$, and $ca = a^2c$ can be represented by commutativity of the following diagrams.


Now we can paste together $10$ of these pentagonal diagrams to form a dodecahedron with two missing faces:

enter image description here

Tracing around the missing faces gives the relation $c = b^{-1}ab^{-1}a^{-3}b$, so $c\in \langle a,b\rangle_G$.


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